What is the integral from 0 to pi/2 of xsin^2(x/2) ?

The integral from 0 to pi/2 of xsin^2(x/2) is (pi^2-4pi+8)/(16)

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integral from 0 to pi/2 of xsin^2(x/2)

\int_{0}^{\frac{π}{2}} x sin^{2} (\frac{x}{2}) d x

\frac{π ^{2} - 4 π + 8}{16}

Decimal

0.33145 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x sin^{2} (\frac{x}{2}) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1}{2} x (1 - cos (x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} x (1 - cos (x)) d x

Apply Integration By Parts

: [x (x - sin (x)) - \int x - sin (x) d x]_{0}^{\frac{π}{2}}

= \frac{1}{2} [x (x - sin (x)) - \int x - sin (x) d x]_{0}^{\frac{π}{2}}

\int x - sin (x) d x = \frac{x ^{2}}{2} + cos (x)

= \frac{1}{2} [x (x - sin (x)) - (\frac{x ^{2}}{2} + cos (x))]_{0}^{\frac{π}{2}}

Simplify

\frac{1}{2} [x (x - sin (x)) - (\frac{x ^{2}}{2} + cos (x))]_{0}^{\frac{π}{2}} : \frac{1}{2} [\frac{x ^{2} - 2 x sin ( x ) - 2 cos ( x )}{2}]_{0}^{\frac{π}{2}}

= \frac{1}{2} [\frac{x ^{2} - 2 x sin ( x ) - 2 cos ( x )}{2}]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π ^{2} - 4 π}{8} + 1

= \frac{1}{2} (\frac{π ^{2} - 4 π}{8} + 1)

Simplify

= \frac{π ^{2} - 4 π + 8}{16}

\int_{0}^{4 π} sin^{2} (x) d x

\int_{0}^{\infty} \frac{3}{x ^{2}} d x

\int_{1}^{\frac{π}{2}} cos (x) d x

\int_{2}^{\infty} \frac{3}{x ^{2}} d x

\int_{0}^{x} \frac{x ^{2}}{2} d x

What is the integral from 0 to pi/2 of xsin^2(x/2) ?
The integral from 0 to pi/2 of xsin^2(x/2) is (pi^2-4pi+8)/(16)